IMPULSE AND MOMENTUM 243
target and the bullet a common velocity v'. Therefore if d! be the new depth of penetration we have
Fd' = i mv* - J (m + M) v'2, (2)
where M is the mass of the target. Eliminating F between equations (1) and (2) we get
.
m \v
But by the conservation of momentum we have
mv — (m 4- M) vf. . (4)
Therefore eliminating the velocities between equations (3) and (4) we get
M ,
M + m
(5)
It is evident from equation (5) that when the target is free, but very large compared with the bullet, the depth penetrated is about the same as when it is fixed.
PROBLEMS.
1. A particle which weighs 2 ounces describes a circle of 1.5 feet radius on a smooth horizontal table. If it makes one complete revolution in every 3 seconds find the magnitude and direction of the impulse imparted by the force, which keeps the particle in the circle,
(a) in one-quarter of a revolution;
(b) in one-half of a revolution;
(c) in three-quarters of a revolution;
(d) in one complete revolution.
2. Find the expression for the impulse imparted to a particle in describing an arc of a circle with uniform speed.
3. Considering the rate of change of the momentum of a particle which describes a uniform circular motion derive the expression for the central force.
4. If we neglect the resistance of the air to the motion of a projectile what can we state with regard to the components of the momentum in the horizontal and vertical directions?
5. A train which weighs 100 tons runs due south at the rate of one mile a minute. Find the lateral force on the western rails due to the rotation of the earth, while the train passes the line of 30° latitude.
6. At what latitude will the force of the preceding problem be a maximum? Determine its amount.